The Question is

What is the minimum value of $9\sec^2 x + \cos^2 x$ ?

Now, I tried applying AM-GM inequality, and the answer comes out as 6. But sec2x has a minimum value of 1. So the least value should be greater than 9.

Now, We have

$\frac{9\sec^2 x + \cos^2 x}2 \ge \sqrt{9\sec^2 x \cos^2 x} $

$\sec^2x\cos^2x = 1$

Now, RHS becomes 3 , and taking the 2 to the RHS , we get

$9\sec^2x + \cos^2x \ge 6$

  • $\begingroup$ There is no way we can do that if you don't show your work. $\endgroup$ – Michael Grant Apr 1 '15 at 14:01
  • $\begingroup$ Is it fine now? $\endgroup$ – Bayern ROX Apr 1 '15 at 14:09
  • 4
    $\begingroup$ Yes. So this tells you the minimum value is greater than or equal to 6. That is correct, right? It it tells you nothing more than that. $\endgroup$ – Michael Grant Apr 1 '15 at 14:11
  • $\begingroup$ Ohh, so in this case, Ive got to check if the two terms are ever equal for the equality to hold.. I see. $\endgroup$ – Bayern ROX Apr 1 '15 at 14:13
  • $\begingroup$ @user31415 Is your comment an April Fool's joke? :) $\endgroup$ – Erick Wong Apr 1 '15 at 15:58

The AM-GM inequality

$$ \sqrt{ab}\le\frac{a+b}{2} $$ becomes less and less tight as the numbers $a$ and $b$ become further and further apart. Indeed, we only get equality if $a=b$.

In your example, we would get equality in the inequality $$ 9\sec^2x + \cos^2x \ge 6 $$ if $9\sec^2x$ were equal to $\cos^2x$ for some $x$. But that never happens - $9\sec^2x$ is always at least $9$ and $\cos^2x$ is always at most $1$.

What this means is that the theoretical lower bound of $6$ - $$ 9\sec^2x + \cos^2x \ge 6 $$ - is never attained. Indeed, since $9\sec^2x$ and $\cos^2x$ are so far apart, the left hand side is actually always substantially larger than $6$. As you point out, it is actually at least $9$.

What's the moral? Use the AM-GM inequality if all you want to know is that one thing is greater than or equal to another. If you're interested in a minimum value that is actually attained, then you'd better make sure that the values you use in the inequality are close together. Otherwise, you'd do better to use another method.

In this case, you're better off using specific analytic properties of the $\cos$ and $\sec$ functions. For example, you might differentiate the expression $\cos^2x+9\sec^2x$. Or you could try to plot the curve and see where it attains its minimum value.

  • $\begingroup$ I see what you mean. Thanks a lot guys $\endgroup$ – Bayern ROX Apr 1 '15 at 14:28

Your inequality is correct, but equality is never reached, so it merely gives you a lower bound. Now if you consider it as a sum of ten terms and use $$\frac{9\sec^2x+\cos^2x}{10}\ge \sqrt[10]{\sec^{18}x\cos^2x}=|\sec x|^{8/5} \ge1$$

equality is possible when $x=0$ hence this would give the minimum. The moral is to use AM-GM while respecting equality conditions.


Hint: the function is periodic with period $\pi$, and you can easely find the minimum value for $x=0$.

  • 2
    $\begingroup$ But I want to know whats wrong in my approach $\endgroup$ – Bayern ROX Apr 1 '15 at 14:09
  • $\begingroup$ Your method says simply that the minimum value is $\ge 6$. And really it is $10$. $\endgroup$ – Emilio Novati Apr 1 '15 at 14:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.